Clearly, $a$ must be non-negative. Furthermore, if $a=0$, the given equation has only one root $x=0$. Hence it must be true that $a > 0$. When $a > 0$, the given equation can be simplified to: $$\frac{x^2}{x-1}=\pm a \Leftrightarrow x^2 - ax + a = 0\quad\text{or}\quad x^2 +ax -a = 0$$ Because $a>0$, the discriminant of the $2^{nd}$ equation is positive which means it has two distinct real roots. It follows that the $1^{st}$ equation should be insolvable in real numbers, or $$a^2-4a < 0 \implies a < 4$$ Hence the final answer is $\boxed{0 < a < 4}$.